The number of silicon atoms per m^{3} is 5 × 10^{28}. This is doped simultaneously with 5 × 10^{22} atoms per m^{3} of Arsenic and 5 × 10^{20} per m^{3} atoms of Indium. Calculate the number of electrons and holes. Given that n_{i}= 1.5 × 10^{16} m^{−3}. Is the material n-type or p-type?

#### Solution

Number of silicon atoms, N = 5 × 10^{28} atoms/m^{3}

Number of arsenic atoms, n_{As} = 5 × 10^{22} atoms/m^{3}

Number of indium atoms, n_{In} = 5 × 10^{20} atoms/m^{3}

Number of thermally-generated electrons, n_{i} = 1.5 × 10^{16} electrons/m^{3}

Number of electrons, n_{e} = 5 × 10^{22} − 1.5 × 10^{16} ≈ 4.99 × 10^{22}

Number of holes = n_{h}

In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:

n_{e}n_{h} = n_{i}^{2}

`therefore "n"_"h" = ("n"_"i"^2)/"n"_"e"`

`= (1.5 xx 10^16)^2/(4.99 xx 10^22) ~~ 4.51 xx 10^9`

Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.